# Determination of the Water Potential of Potato Tuber Cells

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DETERMINATION OF THE WATER POTENTIAL OF POTATO TUBER CELLS.

Method.

Five sucrose solutions with varying molarity and one control containing distilled water were prepared and poured into test tubes. The potato discs were dried, weighed and added to the test tubes. The discs were then weighed again after a period of 24 hours. The percentage change in mass was then calculated.

Apparatus.

&#61623; Specimen tubes with stoppers x6

&#61623; 1cm3 diameter cork borer

&#61623; filter papers

&#61623; balance

&#61623; distilled water

&#61623; sucrose solutions with varying concentrations

&#61623; potato cut into small discs

Results. (Fig 1.0)

Sucrose concentration (M) Initial mass

(kg) Final mass

(kg) Change in mass

(%)

0.10 0.95 1.25 31.57

0.20 0.94 1.13 20.21

0.30 1.03 1.15 11.65

0.40 0.95 1.05 10.52

0.50 0.88 0.86 -2.27

0.60 0.93 0.84 -9.67

Control 0.99 1.40 42.41

Discussion.

Osmosis is the passive diffusion of water molecules across a selectively permeable membrane from a down a concentration gradient. The water potential of a system is the tendency for water to exit the system. In this experiment the aim was to measure the tendency for water to leave the tuber cells. As the water potential of pure water is zero the concentration of sucrose in solution will have an effect on the water potential, this is called the solute potential. The greater the concentration of sucrose the more negative the water potential, because water moves from a high to low water potential.

When the potato is put into water it contains solute molecules which draw water in providing the external solute concentration is lower. The more solute molecules present the lower the water potential such change is referred to as the solute potential. To find the water potential of the cells we need to find out at which concentration of sucrose solutions was a state of equilibrium obtained, i.e. the solution in which there was no change in the volume or mass of the tissue; which is equal to that of the cells. When this happens we know that the solute potential and water potential of the potato and the sucrose solution will be the same. We know this because if this weren't the case then there would be a change in mass.

Figure 1.1 shows the relationship between the molarity of the solution and the percentage change in mass of the tuber sample. This was used to find the concentration at which there was no net exchange of water. I found this to be 0.46mol.dm-3. When I applied this value to figure 1.2 (which displays water potential against sucrose concentration) I found the water potential for 0.48 to be -1420 kPa, which as explained is the same as the water potential for the potato tuber cells.

The graph plotted in figure 1.1 showed the kind of downward curve I expected to see. At 0.0M the % change in mass is 42.41%, what's happened here is the cells have taken up a lot of water and become almost fully turgid. This happens because their water potential is more negative than that of distilled water, which is zero kPa. When a cell becomes fully turgid the pressure exerted on the cell wall prevents anymore water from entering. This technique can be observed in guard cells undertaking stomatal opening. A stoma opens when its guard cells actively take up K+ and water from surrounding cells enters by osmosis. When the vacuoles in the guard cells gain water, the cells become turgid and swell. This causes them to buckle outward increasing the size of the gap (stomatal aperture) between them. When the stoma is open the plant takes in CO2 during the day for photosynthesis. However turgidity is not only used in stomatal activity. It is an

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